Find the distance between the parallel lines | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The distance $(d)$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula: $d = \frac{|C_1 - C_2|}{\sqrt{A

Vedclass · Accepted Answer

$65/17$

Vedclass · Answer

(C) The distance $(d)$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula: $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$ Given the lines: $15x + 8y - 34 = 0$ $15x + 8y + 31 = 0$ Here,$A = 15$,$B = 8$,$C_1 = -34$,and $C_2 = 31$. Substituting these values into the formula: $d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}}$ $d = \frac{|-65|}{\sqrt{225 + 64}}$ $d = \frac{65}{\sqrt{289}}$ $d = \frac{65}{17}$ units.

Find the distance between the parallel lines $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$.

Similar Questions

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

The length of the perpendicular drawn from the origin $(0, 0)$ to the line joining the points $(x', y')$ and $(x'', y'')$ is:

Find the length of the perpendicular drawn from the point $(2, 1)$ to the line $3x - 4y + 8 = 0$.

The distance between the two parallel lines $y = 2x + 7$ and $y = 2x + 5$ is:

If $(\sin \theta, \cos \theta)$ and $(3, 2)$ lie on the same side of the line $x + y = 1$,then $\theta$ lies in the interval:

Vedclass Test Series

Exam Paper Generator

Online Exam Module